 Solution

False. Take $\kappa ={\aleph }_0$, $\lambda =1$, and ${\lambda }^{\prime }=2$. ThenFor example the Cantor pairing function $⟨m,n⟩=(m+n)(m+n+1)/2+n$ bijects ${\mathbb{N}}^2$ with $\mathbb{N}$. Thus the exponent increases strictly while the cardinal power remains the same. This counterexample is completely choice-free.